# Differentiation under the integral sign

Differentiation under the integral sign is a useful operation in calculus. Formally it can be stated as follows:

Theorem. Let f(x, t) be a function such that both f(x, t) and its partial derivative fx(x, t) are continuous in t and x in some region of the (x, t)-plane, including a(x) ≤ tb(x), x0xx1. Also suppose that the functions a(x) and b(x) are both continuous and both have continuous derivatives for x0xx1. Then for x0xx1:
$\frac{\mathrm{d}}{\mathrm{d}x} \left (\int_{a(x)}^{b(x)}f(x,t)\,\mathrm{d}t \right) = f(x,b(x))\cdot b'(x) - f(x,a(x))\cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x}f(x,t)\; \mathrm{d}t.$

This formula is the general form of the Leibniz integral rule and can be derived using the fundamental theorem of calculus. The [second] fundamental theorem of calculus is just a particular case of the above formula, for a(x) = a, a constant, b(x) = x and f(x, t) = f(t).

If both upper and lower limits are taken as constants, then the formula takes the shape of an operator equation:

ItDx = DxIt,

where Dx is the partial derivative with respect to x and It is the integral operator with respect to t over a fixed interval. That is, it is related to the symmetry of second derivatives, but involving integrals as well as derivatives. This case is also known as the Leibniz integral rule.

The following three basic theorems on the interchange of limits are essentially equivalent:

• the interchange of a derivative and an integral (differentiation under the integral sign; i.e., Leibniz integral rule)
• the change of order of partial derivatives
• the change of order of integration (integration under the integral sign; i.e., Fubini's theorem)

## Higher dimensions

The Leibniz integral rule can be extended to multidimensional integrals. In two and three dimensions, this rule is better known from the field of fluid dynamics as the Reynolds transport theorem:

$\frac{\mathrm{d}}{\mathrm{d}t} \int_{D(t)} F(\vec{\textbf x}, t) \,\mathrm{d}V = \int_{D(t)} \frac{\partial}{\partial t} \,F(\vec{\textbf x}, t)\,\mathrm{d}V + \int_{\partial D(t)} \,F(\vec{\textbf x}, t)\, \vec{\textbf v}_b \cdot \mathrm{d}\mathbf{\Sigma}$

where $F(\vec{\textbf x}, t)\,$ is a scalar function, D(t) and ∂D(t) denote a time-varying connected region of R3 and its boundary, respectively, $\vec{\textbf v}_b\,$ is the Eulerian velocity of the boundary (see Lagrangian and Eulerian coordinates) and dΣ = n dS is the unit normal component of the surface element.

The general statement of the Leibniz integral rule requires concepts from differential geometry, specifically differential forms, exterior derivatives, wedge products and interior products. With those tools, the Leibniz integral rule in p-dimensions is:[1]

$\frac{\mathrm{d}}{\mathrm{d}t}\int_{\Omega(t)}\omega=\int_{\Omega(t)} i_{\vec{\textbf v}}(\mathrm{d}_x\omega)+\int_{\partial \Omega(t)} i_{\vec{\textbf v}} \omega+\int_{\Omega(t)}\dot{\omega},\,$

where Ω(t) is a time-varying domain of integration, ω is a p-form, $\vec{\textbf v}\,$ is the vector field of the velocity, $\vec{\textbf v}=\frac{\partial\vec{\textbf x}}{\partial t}\,$, i denotes the interior product, dxω is the exterior derivative of ω with respect to the space variables only and $\dot{\omega}\,$ is the time-derivative of ω.

## Proof of Theorem

Lemma. One has:
$\frac{\partial}{\partial b} \left (\int_a^b f(x)\; \mathrm{d}x \right ) = f(b), \qquad \frac{\partial}{\partial a} \left (\int_a^b f(x)\; \mathrm{d}x \right )= -f(a).$

Proof. From proof of the fundamental theorem of calculus,

\begin{align} \frac{\partial}{\partial b} \left (\int_a^b f(x)\; \mathrm{d}x \right ) &= \lim_{\Delta b \to 0} \frac{1}{\Delta b} \left[ \int_a^{b+\Delta b} f(x)\,\mathrm{d}x - \int_a^b f(x)\,\mathrm{d}x \right] \\ &= \lim_{\Delta b \to 0} \frac{1}{\Delta b} \int_b^{b+\Delta b} f(x)\,\mathrm{d}x \\ &= \lim_{\Delta b \to 0} \frac{1}{\Delta b} \left[ f(b) \Delta b + \mathcal{O}\left(\Delta b^2\right) \right] \\ &= f(b) \\ \frac{\partial}{\partial a} \left (\int_a^b f(x)\; \mathrm{d}x \right )&= \lim_{\Delta a \to 0} \frac{1}{\Delta a} \left[ \int_{a+\Delta a}^b f(x)\,\mathrm{d}x - \int_a^b f(x)\,\mathrm{d}x \right] \\ &= \lim_{\Delta a \to 0} \frac{1}{\Delta a} \int_{a+\Delta a}^a f(x)\,\mathrm{d}x \\ &= \lim_{\Delta a \to 0} \frac{1}{\Delta a} \left[ -f(a)\, \Delta a + \mathcal{O}\left(\Delta a^2\right) \right]\\ &= -f(a). \end{align}

Suppose a and b are constant, and that f(x) involves a parameter α which is constant in the integration but may vary to form different integrals. Assuming that f(x, α) is a continuous function of x and α in the compact set {(x, α) : α0 ≤ α ≤ α1 and axb}, and that the partial derivative fα(x, α) exists and is continuous, then if one defines:

$\varphi(\alpha) = \int_a^b f(x,\alpha)\;\mathrm{d}x.$
$\varphi$ may be differentiated with respect to α by differentiating under the integral sign; i.e.,
$\frac{\mathrm{d}\varphi}{\mathrm{d}\alpha}=\int_a^b\frac{\partial}{\partial\alpha}\,f(x,\alpha)\,\mathrm{d}x.\,$

By the Heine–Cantor theorem it is uniformly continuous in that set. In other words for any ε > 0 there exists Δα such that for all values of x in [a, b]:

$|f(x,\alpha+\Delta \alpha)-f(x,\alpha)|<\varepsilon.$

On the other hand:

\begin{align} \Delta\varphi &=\varphi(\alpha+\Delta \alpha)-\varphi(\alpha) \\ &=\int_a^b f(x,\alpha+\Delta\alpha)\;\mathrm{d}x - \int_a^b f(x,\alpha)\; \mathrm{d}x \\ &=\int_a^b \left (f(x,\alpha+\Delta\alpha)-f(x,\alpha) \right )\;\mathrm{d}x \\ &\leq \varepsilon (b-a) \end{align}

Hence φ(α) is a continuous function.

Similarly if $\frac{\partial}{\partial\alpha}\,f(x,\alpha)$ exists and is continuous, then for all ε > 0 there exists Δα such that:

$\forall x \in [a, b] \quad \left|\frac{f(x,\alpha+\Delta \alpha)-f(x,\alpha)}{\Delta \alpha} - \frac{\partial f}{\partial\alpha}\right|<\varepsilon.$

Therefore,

$\frac{\Delta \varphi}{\Delta \alpha}=\int_a^b\frac{f(x,\alpha+\Delta\alpha)-f(x,\alpha)}{\Delta \alpha}\;\mathrm{d}x = \int_a^b \frac{\partial\,f(x,\alpha)}{\partial \alpha}\,\mathrm{d}x + R$

where

$|R| < \int_a^b \varepsilon\; \mathrm{d}x = \varepsilon(b-a).$

Now, ε → 0 as Δα → 0, therefore,

$\lim_{{\Delta \alpha} \rarr 0}\frac{\Delta\varphi}{\Delta \alpha}= \frac{\mathrm{d}\varphi}{\mathrm{d}\alpha} = \int_a^b \frac{\partial}{\partial \alpha}\,f(x,\alpha)\,\mathrm{d}x.\,$

This is the formula we set out to prove.

Now, suppose

$\int_a^b f(x,\alpha)\;\mathrm{d}x=\varphi(\alpha),$

where a and b are functions of α which take increments Δa and Δb, respectively, when α is increased by Δα. Then,

\begin{align} \Delta\varphi &=\varphi(\alpha+\Delta\alpha)-\varphi(\alpha) \\ &=\int_{a+\Delta a}^{b+\Delta b}f(x,\alpha+\Delta\alpha)\;\mathrm{d}x\,-\int_a^b f(x,\alpha)\;\mathrm{d}x\, \\ &=\int_{a+\Delta a}^af(x,\alpha+\Delta\alpha)\;\mathrm{d}x+\int_a^bf(x,\alpha+\Delta\alpha)\;\mathrm{d}x+\int_b^{b+\Delta b}f(x,\alpha+\Delta\alpha)\;\mathrm{d}x -\int_a^b f(x,\alpha)\;\mathrm{d}x \\ &=-\int_a^{a+\Delta a}\,f(x,\alpha+\Delta\alpha)\;\mathrm{d}x+\int_a^b[f(x,\alpha+\Delta\alpha)-f(x,\alpha)]\;\mathrm{d}x+\int_b^{b+\Delta b}\,f(x,\alpha+\Delta\alpha)\;\mathrm{d}x. \end{align}

A form of the mean value theorem, $\int_a^bf(x)\;\mathrm{d}x=(b-a)f(\xi),$ where a < ξ < b, can be applied to the first and last integrals of the formula for Δφ above, resulting in

$\Delta\varphi=-\Delta a\,f(\xi_1,\alpha+\Delta\alpha)+\int_a^b[f(x,\alpha+\Delta\alpha)-f(x,\alpha)]\;\mathrm{d}x+\Delta b\,f(\xi_2,\alpha+\Delta\alpha).$

Dividing by Δα, letting Δα → 0, noticing ξ1a and ξ2b and using the above derivation for

$\frac{\mathrm{d}\varphi}{\mathrm{d}\alpha} = \int_a^b\frac{\partial}{\partial \alpha}\,f(x,\alpha)\,\mathrm{d}x$

yields

$\frac{\mathrm{d}\varphi}{\mathrm{d}\alpha} = \int_a^b\frac{\partial}{\partial \alpha}\,f(x,\alpha)\,\mathrm{d}x+f(b,\alpha)\frac{\partial b}{\partial \alpha}-f(a,\alpha)\frac{\partial a}{\partial \alpha}.$

This is the general form of the Leibniz integral rule.

## Examples

### General examples

#### Example 1

$\varphi(\alpha)=\int_0^1\frac{\alpha}{x^2+\alpha^2}\;\mathrm{d}x = \begin{cases} 0 & \alpha = 0 \\ \arctan \left ( \tfrac{1}{\alpha} \right ) & \alpha \neq 0 \end{cases}$

The function under the integral sign is not continuous at the point (x, α) = (0, 0) and the function φ(α) has a discontinuity at α = 0, because φ(α) approaches ±π/2 as α → 0±.

If we now differentiate φ(α) with respect to α under the integral sign, we get

$\frac{\mathrm{d}}{\mathrm{d}\alpha} \varphi(\alpha)=\int_0^1\frac{\partial}{\partial\alpha}\left(\frac{\alpha}{x^2+\alpha^2}\right)\;\mathrm{d}x=\int_0^1\frac{x^2-\alpha^2}{(x^2+\alpha^2)^2} \mathrm{d}x=-\frac{x}{x^2+\alpha^2}\bigg|_0^1=-\frac{1}{1+\alpha^2},$

which is, of course, true for all values of α except α = 0.

#### Example 2

An example with variable limits:

\begin{align} \frac{\mathrm{d}}{\mathrm{d}x} \int_{\sin x}^{\cos x} \cosh t^2\;\mathrm{d}t &= \cosh\left(\cos^2 x\right) \frac{\mathrm{d}}{\mathrm{d}x}(\cos x) - \cosh\left(\sin^2 x\right) \frac{\mathrm{d}}{\mathrm{d}x} (\sin x) + \int_{\sin x}^{\cos x} \frac{\partial}{\partial x} \left (\cosh t^2\right )\mathrm{d}t \\ &= \cosh\left(\cos^2 x\right) (-\sin x) - \cosh\left(\sin^2 x\right) (\cos x) + 0 \\ &= - \cosh\left(\cos^2 x\right) \sin x - \cosh\left(\sin^2 x\right) \cos x \end{align}

### Examples for evaluating a definite integral

#### Example 3

The principle of differentiating under the integral sign may sometimes be used to evaluate a definite integral. Consider:

$\varphi(\alpha)=\int_0^\pi\,\ln(1-2\alpha\cos(x)+\alpha^2)\;\mathrm{d}x \qquad |\alpha| > 1.$

Now,

\begin{align} \frac{\mathrm{d}}{\mathrm{d}\alpha}\,\varphi(\alpha) &=\int_0^\pi \frac{-2\cos(x)+2\alpha }{1-2\alpha \cos(x)+\alpha^2}\;\mathrm{d}x\, \\[8pt] &=\frac{1}{\alpha}\int_0^\pi\,\left(1-\frac{1-\alpha^2}{1-2\alpha \cos(x)+\alpha^2}\,\right)\,\mathrm{d}x \\[8pt] &=\frac{\pi}{\alpha}-\frac{2}{\alpha}\left\{\,\arctan\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,\right\}\,\bigg|_0^\pi. \end{align}

Now as x varies from 0 to π we have:

$\begin{cases} \frac{1+\alpha}{1-\alpha} \tan\left(\frac{x}{2}\right) \geq 0 & |\alpha| < 1 \\ \frac{1+\alpha}{1-\alpha} \tan \left( \frac{x}{2}\right) \leq 0 & |\alpha| > 1 \end{cases}$

Hence,

$\arctan\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,\bigg|_0^\pi= \begin{cases} \frac{\pi}{2} & |\alpha| < 1 \\ -\frac{\pi}{2} & |\alpha| > 1 \end{cases}$

Therefore,

$\frac{\mathrm{d}}{\mathrm{d}\alpha}\,\varphi(\alpha)= \begin{cases} 0 & |\alpha| < 1 \\ \frac{2\pi}{\alpha} & |\alpha| > 1 \end{cases}$

Integrating both sides with respect to α, we get:

$\varphi (\alpha) = \begin{cases} C_1 & |\alpha| < 1 \\ 2\pi \ln |\alpha| + C_2 & |\alpha| > 1 \end{cases}$

C1 = 0 follows from evaluating φ(0):

$\varphi(0) =\int_0^\pi \ln(1)\;\mathrm{d}x =\int_0^\pi 0\;\mathrm{d}x=0$

To determine C2 in the same manner, we should need to substitute in a value of α greater than 1 in φ(α). This is somewhat inconvenient. Instead, we substitute α = 1/β, where |β| < 1. Then,

\begin{align} \varphi(\alpha) &=\int_0^\pi\left(\ln(1-2\beta \cos(x)+\beta^2)-2\ln|\beta|\right)\;\mathrm{d}x\ \\[8pt] &=0-2\pi\ln|\beta|\, \\[8pt] &=2\pi\ln|\alpha|\, \end{align}

Therefore, C2 = 0.

The definition of φ(α) is now complete:

$\varphi (\alpha) = \begin{cases} 0 & |\alpha| < 1 \\ 2\pi \ln |\alpha| & |\alpha| > 1 \end{cases}$

The foregoing discussion, of course, does not apply when α = ±1, since the conditions for differentiability are not met.

#### Example 4

$\textbf I = \int_0^{\frac{\pi}{2}}\frac{1}{\left(a\cos^2(x)+b\sin^2 (x)\right)^2}\;\mathrm{d}x,\qquad a,b > 0.$

First we calculate:

\begin{align} \textbf{J} &= \int_0^{\frac{\pi}{2}} \frac{1}{a\cos^2 (x) + b \sin^2 (x)}\;\mathrm{d}x \\ [6pt] &= \int_0^{\frac{\pi}{2}} \frac{\frac{1}{\cos^2 (x)}}{a + b \frac{\sin^2 (x)}{\cos^2 (x)}} \mathrm{d}x \\[6pt] &= \int_0^{\frac{\pi}{2}} \frac{\sec^2(x)}{a +b \tan^2(x)}\;\mathrm{d}x \\ [6pt] &= \frac{1}{b} \int_0^{\frac{\pi}{2}} \frac{1}{\left(\sqrt{\frac{a}{b}}\right)^2+\tan^2 (x)}\;\mathrm{d}(\tan x) \\[6pt] &= \frac{1}{\sqrt{ab}}\arctan \left(\sqrt{\frac{b}{a}}\tan (x)\right) \Bigg|_0^{\frac{\pi}{2}} \\[6pt] &=\frac{\pi}{2\sqrt{ab}}. \end{align}

The limits of integration being independent of a, But we have:

$\frac{\partial \textbf J}{\partial a}=-\int_0^{\frac{\pi}{2}} \frac{\cos^2 x\;\mathrm{d}x}{\left(a\cos^2 x+b \sin^2 x\right)^2}$

On the other hand:

$\frac{\partial \textbf J}{\partial a}= \frac{\partial}{\partial a} \left(\frac{\pi}{2\sqrt{ab}}\right) =-\frac{\pi}{4\sqrt{a^3b}}.$

Equating these two relations then yields

$\int_0^{\frac{\pi}{2}} \frac{\cos^2(x)\;\mathrm{d}x}{\left(a \cos^2(x)+b \sin^2(x)\right)^2}=\frac{\pi}{4\sqrt{a^3b}}.$

In a similar fashion, pursuing $\frac{\partial \textbf J}{\partial b}$ yields

$\int_0^{\frac{\pi}{2}}\frac{\sin^2 x \mathrm{d}x}{\left(a\cos^2 x+b\sin^2 x\right)^2} = \frac{\pi}{4\sqrt{ab^3}}.$

Adding the two results then produces

$\textbf I = \int_0^{\frac{\pi}{2}}\frac{1}{\left(a\cos^2x+b\sin^2x\right)^2}\;\mathrm{d}x=\frac{\pi}{4\sqrt{ab}}\left(\frac{1}{a}+\frac{1}{b}\right),$

Note that if we define

$\textbf I_n = \int_0^{\frac{\pi}{2}} \frac{1}{\left(a\cos^2 x+b\sin^2 x\right)^n}\;\mathrm{d}x,$

it can easily be shown that

$\frac{\partial\textbf I_{n-1}}{\partial a} + \frac{\partial \textbf I_{n-1}}{\partial b} + (n-1)\textbf I_n = 0.$

Given I1, this partial-derivative-based recursive relation (i.e., integral reduction formula) can then be utilized to compute all of the values of In for n > 1.

#### Example 5

Here, we consider the integral

$\textbf I(\alpha)=\int_0^{\frac{\pi}{2}}\frac{\ln\,(1+\cos\alpha\,\cos\,x)}{\cos x}\;\mathrm{d}x, \qquad 0 < \alpha < \pi.$

Differentiating under the integral with respect to α, we have

\begin{align} \frac{\mathrm{d}}{\mathrm{d}\alpha} \textbf{I}(\alpha) &= \int_0^{\frac{\pi}{2}} \frac{\partial}{\partial\alpha} \left(\frac{\ln(1 + \cos\alpha \cos x)}{\cos x}\right)\,\mathrm{d}x \\ [6pt] &=-\int_0^{\frac{\pi}{2}}\frac{\sin \alpha}{1+\cos \alpha \cos x}\,\mathrm{d}x \\ [6pt] &=-\int_0^{\frac{\pi}{2}}\frac{\sin \alpha}{\left(\cos^2 \frac{x}{2}+\sin^2 \frac{x}{2}\right)+\cos \alpha \left(\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}\right)}\,\mathrm{d}x \\ [6pt] &=-\frac{\sin\alpha}{1-\cos\alpha} \int_0^{\frac{\pi}{2}} \frac{1}{\cos^2\frac{x}{2}} \frac{1}{\frac{1+\cos \alpha}{1-\cos \alpha} +\tan^2 \frac{x}{2}}\,\mathrm{d}x \\ [6pt] &=-\frac{2\sin\alpha}{1-\cos\alpha} \int_0^{\frac{\pi}{2}} \frac{\frac{1}{2} \sec^2 \frac{x}{2}}{\frac{2 \cos^2 \frac{\alpha}{2}}{2 \sin^2\frac{\alpha}{2}} + \tan^2 \frac{x}{2}} \,\mathrm{d}x \\ [6pt] &=-\frac{2\left(2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}\right)}{2 \sin^2 \frac{\alpha}{2}} \int_0^{\frac{\pi}{2}}\,\frac{1}{\cot^2\frac{\alpha}{2} + \tan^2 \frac{x}{2}} \mathrm{d}\left(\tan \frac{x}{2}\right)\\ [6pt] &=-2\cot \frac{\alpha}{2}\int_0^{\frac{\pi}{2}} \frac{1}{\cot^2\frac{\alpha}{2} + \tan^2\frac{x}{2}}\,\mathrm{d}\left(\tan \frac{x}{2}\right)\\ [6pt] &=-2\arctan \left(\tan \frac{\alpha}{2} \tan \frac{x}{2} \right) \bigg|_0^{\frac{\pi}{2}}\\ &=-\alpha \end{align}

Therefore:

$\textbf{I}(\alpha) = C - \frac{\alpha^2}{2}$

However by definition, I(π/2) = 0, hence: C = π2/8 and

$\textbf I(\alpha) = \frac{\pi^2}{8}-\frac{\alpha^2}{2}.$

#### Example 6

Here, we consider the integral

$\int_0^{2\pi}e^{\cos\theta} \cos(\sin\theta)\;\mathrm{d}\theta.$

We introduce a new variable φ and rewrite the integral as

$f(\varphi) = \int_0^{2\pi} e^{\varphi\cos\theta} \cos(\varphi\sin\theta)\;\mathrm{d}\theta.$

Note that for φ = 1 we recover the original integral, now we proceed:

\begin{align} \frac {\mathrm{d}f}{\mathrm{d}\varphi} &= \int_0^{2\pi} \frac{\partial}{\partial\varphi}\left(e^{\varphi\cos\theta}\;\cos(\varphi\sin\theta)\right)\;\mathrm{d}\theta \\ &= \int_0^{2\pi} e^{\varphi\cos\theta} \left(\cos\theta\cos(\varphi\sin\theta)-\sin\theta\sin(\varphi\sin\theta)\right)\;\mathrm{d}\theta \\ &= \int_0^{2\pi} \frac {1}{\varphi}\;\frac {\partial}{\partial\theta}\left(e^{\varphi\cos\theta} \sin(\varphi\sin\theta)\right)\;\mathrm{d}\theta \\ &= \frac {1}{\varphi} \int_0^{2\pi}\;\mathrm{d}\left(e^{\varphi\cos\theta} \sin(\varphi\sin\theta)\right) \\ &= \frac {1}{\varphi} \left(e^{\varphi\cos\theta}\;\sin(\varphi\sin\theta)\right)\;\bigg|_0^{2\pi} = 0. \end{align}

Integrating both sides of $\frac {\mathrm{d}f}{\mathrm{d}\varphi}=0$ with respect to φ between the limits 0 and 1 yields

$f(1) - f(0) = \int_{f(0)}^{f(1)}\;\mathrm{d}f = \int_{0}^1 0\;\mathrm{d}\varphi = 0$

Therefore f(1) = f(0) however we note that from the equation for f(φ), we have f(0) = 2π, therefore the value of f at φ = 1, which is the same as the integral we set out to compute is 2π.

#### Other problems to solve

There are innumerable other integrals that can be solved "quickly" using the technique of differentiation under the integral sign. For example consider the following cases where one adds a new variable α:

\begin{align} \int_0^\infty\;\frac{\sin\,x}{x}\;\mathrm{d}x &\to \int_0^\infty\;e^{-\alpha\,x}\;\frac{\sin\,x}{x}\;\mathrm{d}x,\\ \int_0^{\frac{\pi}{2}}\;\frac{x}{\tan\,x}\;\mathrm{d}x &\to\int_0^{\frac{\pi}{2}}\;\frac{\tan^{-1}(\alpha\,\tan\,x)}{\tan\,x}\;\mathrm{d}x,\\ \int_0^{\infty}\;\frac{\ln\,(1+x^2)}{1+x^2}\;\mathrm{d}x &\to\int_0^{\infty}\;\frac{\ln\,(1+\alpha^2\,x^2)}{1+x^2}\;\mathrm{d}x \\ \int_0^1\;\frac{x-1}{\ln\,x}\;\mathrm{d}x &\to \int_0^1\;\frac{x^\alpha-1}{\ln\,x}\;\mathrm{d}x. \end{align}

The first integral, the Dirichlet integral, is absolutely convergent for positive α but only conditionally convergent when α is 0. Therefore differentiation under the integral sign is easy to justify when α > 0, but proving that the resulting formula remains valid when α is 0 requires some careful work.

### Applications to series

Differentiating under the integral can also be applied to differentiating under summation, interpreting summation as counting measure. An example of an application is the fact that power series are differentiable in their radius of convergence.

## Popular culture

Differentiation under the integral sign is mentioned in the late physicist Richard Feynman's best-selling memoir Surely You're Joking, Mr. Feynman! (in the chapter "A Different Box of Tools"), where he mentions learning it from an old text, Advanced Calculus (1926), by Frederick S. Woods (who was a professor of mathematics in the Massachusetts Institute of Technology) while in high school. The technique was not often taught when Feynman later received his formal education in calculus and, knowing it, Feynman was able to use the technique to solve some otherwise difficult integration problems upon his arrival at graduate school at Princeton University. The direct quotation from Surely You're Joking, Mr. Feynman! regarding the method of differentiation under the integral sign is as follows: